A generalization of Liouville's theorem

Question

Let $A\subseteq \Bbb{C}$ be countable and closed. Let $f: \Bbb{C}\backslash A \longrightarrow \Bbb{C}$ be analytic and bounded. Is it true that $f$ must be constant ?

Answer 1

The proof proposed by edo arad is quite nice.

Here is another one, which is essentially the same but avoids the use of transfinite induction.

By Liouville, it is enough to show that $f$ can be extended to an analytic function on $\mathbb C$.

To do this, it is enough to show that every point $a\in A$ has an open neighbourhood $V_a$ such that $f$ can be extended to an analytic function $f_a$ on $(\mathbb C\setminus A)\cup V_a$. Indeed, assume that this holds true. Then the $V_a$'s can be taken to be open disks, so that $V_a\cup V_{a'}$ is connected whenever $V_a\cap V_{a'}\neq\emptyset$. By the identity principle, it follows that if $V_a\cap V_{a'}\neq\emptyset$, then $f_a\equiv f_{a'}$ on $V_a\cap V_{a'}$: this is because $f_a$ and $f_{a'}$ agree on $V_a\cap V_{a'}\cap (\mathbb C\setminus A)$, which is a nonempty open set because $\mathbb C\setminus A$ is dense in $\mathbb C$ (recall that $A$ is countable, so it has empty interior). Therefore, there is a perfectly defined function $\widetilde f:\mathbb C\to\mathbb C$ such that $\widetilde f\equiv f$ on $\mathbb C\setminus A$ and $\widetilde f(z)\equiv f_a(z)$ on $V_a$ for any $a\in A$; and $\widetilde f$ is analytic because it is analytic in a neighbourhood of each point.

Let us denote by $S$ the set of all points $a\in S$ such that $f$ cannot be analytically extended in a neighbourhood of $a$. We have to show that $S=\emptyset$; and since $S$ is countable and obviously closed, it is enough to show that $S$ has no isolated points (because a nonempty closed subset of $\mathbb C$ without isolated points has to be uncountable, for example by the Baire category theorem).

Towards a contradiction, assume that $S$ has at least one isolated point $a$. This means that one can find an open neighbourhood $V$ of $a$ such that $V\cap S=\{ a\}$. In other words, $f$ can be analytically extended in a neighbourhood of every point $a'\in V\setminus\{ a\}$. By the above reasoning, $f_{V\setminus A}$ can be extended to an analytic function $g$ defined on $V\setminus\{ a\}$. This function $g$ is bounded on $V\setminus\{ a\}$ because $f$ is bounded on $V\setminus A$ and $V\setminus A$ is dense in $V$. But this implies that $g$ has a removable singularity at $a$. So $f_{V\setminus A}$ can in fact be extended to an analytic function on $V$, which contradicts the assumption $a\in S$.

Answer 2

We prove that $f$ must be constant:

First, if $A$ is a single point then $f$ has a removable singularity on $A$, so we can extend $f$ holomorphically to $\mathbb{C}$ and use Liouville. Similarly for the case of $|A|$ finite.

Now we want to reduce to the case where $A$ does not have any isolated points, and for this you can use use transfinite induction (please let me know if you find a way to bypass it). We define the set $A_\alpha$ (for $\alpha$ an ordinal) to be

• $A$, if $\alpha=0$
• $A_{\alpha-1}-{p_\alpha}$ where $p_\alpha$ is an isolated point, if $\alpha$ is a successor and there exist such $p_\alpha\in A_{\alpha-1}$.
• $\cap_{\beta<\alpha}A_\beta-p_\alpha$ if $\alpha$ is a limit and there exist such $p_\alpha\in A_{\alpha-1}$.
• $A_{\alpha-1}$ if $\alpha$ if $\alpha$ a successor and there are no isolated points left in $A_{\alpha-1}$.
• $\cap_{\beta<\alpha}A_\beta$ if $\alpha$ is a limit and there are no isolated points in $\cap_{\beta<\alpha}A_\beta$.

Now notice that for some $\alpha_0$ this sequence will stabilize ($A_{\alpha_0}=A_{{\alpha_0}+1}=...$), since $A$ is countable and the set $A-A_\alpha$ is of the same cardinality as $\alpha$ (and there are uncountable ordinals). Now we prove that for any $\alpha$ we can define $f$ on $A_\alpha$:

For $\alpha=0$ it is clear. for $\alpha$ successor, the point $p_\alpha$ is a removable singularity for $f$ extended to $A_{\alpha-1}$, so we can extend $f$ further to $A_\alpha$. For $\alpha$ a limit ordinal, the point $p_\alpha$ is a removable singularity for $f$ extended to $cap_{\beta<\alpha}A_\beta$, so we can extend $f$ further to $A_\alpha$. (notice that both here and in the construction above we could have written the successor step exactly the same as the limit step).

So we conclude that we can extend $f$ to $A_{\alpha_0}$ which does not have any isolated points (and of course $f$ remains bounded and analytic). as an exercise, use transfinite induction as above to show that all $A_\alpha$ are closed. So we assume that $A$ does not have any isolated points.

Now we use this question which shows that there are no (non-empty) closed uncountable sets without isolated points, so $A=\emptyset$ and we can directly use Liouville.